Question

If the distance of the point $P(1,–2,1)$ from the plane $x+2y–2z=\alpha$, where $\alpha >0$, is $5$, then the foot of the perpendicular from $P$ to the plane is

• A. $(\frac{2}{3},-\frac{1}{3},\frac{5}{3})$
• B. $(\frac{1}{3},\frac{2}{3},\frac{10}{3})$
• C. $(\frac{4}{3},-\frac{4}{3},\frac{1}{3})$
• D. $(\frac{8}{3},\frac{4}{3},-\frac{7}{3})$

Answer 1

The correct option is B $(\frac{1}{3},\frac{2}{3},\frac{10}{3})$

Distance of point $P$ from plane $=5$
$\therefore 5=\left|\frac{1-4-2-\alpha }{3}\right|$
$\alpha =10$
Foot of perpendicular
$\frac{x-1}{1}=\frac{y+2}{2}=\frac{z-1}{-2}=\frac{5}{3}$
$x=\frac{8}{3},y=\frac{4}{3},z=-\frac{7}{3}$
Thus the foot of the perpendicular is
$\therefore f(\frac{8}{3},\frac{4}{3},-\frac{7}{3})$